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The given operation is done with $6$ significant digits (mantissa) and in base $10$

$$10^6+\underbrace{1+1+\ldots+1}_{10^6\text{-times}} $$

Now we had to get $2*10^6$ but we get $10^6$, why is that?

I understand that the number is stored as:

$\sigma\cdot (m)_{B}\cdot B^{exp}$

So $$10^6=(-1)^0*1.000000*10^6$$

So $$1=(-1)^0*1.000000*10^0\text{ or }1=(-1)^0*0.000001*10^6?$$ Why does it turn out that $10^6+1=10^6?$

gbox
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  • "Now we had to get $2*10^6$ but we get $10^6$, why is that?": why are you expecting to get $2 \times 10^6?$ – Ben Grossmann Jan 31 '18 at 15:58
  • Uh, no.... why had to get $10^6 + n$. Right? – fleablood Jan 31 '18 at 16:10
  • Sorry all copied the mathjax code and did not edit the n-times – gbox Jan 31 '18 at 16:15
  • Well because you are adding $1$ consecutive times. each time you add $1$ it will round down and put you back to $10^6$ so you'll never get anywhere. But if you did $10^6 + 10$ and added $10$ up $10^5$ times you will get $2*10^6$. – fleablood Jan 31 '18 at 16:18
  • If you can't add something once (and $10^6 + 1$ has seven digits so the $1$ is not significant so you can't add it) then you can't add it $10^6$ times. – fleablood Jan 31 '18 at 16:19

1 Answers1

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Perhaps because to $6$ significant digits, $\color{#C00}{100000}1=\color{#C00}{100000}0$ when rounded. If you are really working in base ten, unlike binary, you can't assume the first digit is $1$ and only count the remaining digits for the $6$ significant digits. This would mean that only the parts in red are being retained by the calculator/computer.

Try incrementing by $10$ and see if the additions stick.

robjohn
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